2
$\begingroup$

In a previous question of mine, I asked how the volume of the region $$xy>zw \quad\wedge \quad x>-y$$ bounded by the unit $4$-ball (so $x^2+y^2+z^2+w^2<1$) could be calculated.

The answer to my question implied that the answer ($1/4$ the volume of the unit $4$-ball) could easily be seen through symmetry, but I never really understood the arguments, hence this question.

I'd really appreciate either a new perspective on this, or an explanation of the answerers argument, which roughly goes as follows:

  1. The condition $0<x+y$ divides the circle (the area of the projection of the total region on the $x,y$ plane, given the values of $z$ and $w$) in half (on the $x,y$ plane).
  2. Folding the negative $z$ half plane onto the positive one, to get the complementary superposition, divides the volume in half (on the $z,w$ plane).

The total result is to divide the whole ball by $4$.

I believe I understand the first part, but the second part eludes me; I cannot seem to visualize this "folding".

I think the idea is to say that for every pair of values of $(z,w)$ and $(-z,w),$ the sum of the areas of the projections of the region in question onto the $x,y$ plane for these two pairs ($(z,w)$ and $(-z,w)$) is a quarter of the area of the circle, or something like that. The answerer writes "the sum of the two areas, for the same $|z \cdot w|$ value, totals that of the semicircle." But I can't see why this should be true from the symmetries of the region.

Thank you.

$\endgroup$
1
$\begingroup$

Hint Each point on the $4$-ball is in precisely one of the subsets $$M_+ = \{xy > zw\}, \qquad M_- = \{xy < zw\}, \qquad M_0 = \{xy = zw\}$$ of the $4$-ball, and, being Zariski closed (and not the whole space), $M_0$ does not contribute to the volume of the region. On the other hand, the volume-preserving map $\phi : (x, y, z, w) \mapsto (-x, y, -z, w)$ interchanges $M_+$ and $M_-$.

One can readily modify this argument to treat the condition $x > -y$ simultaneously.

$\endgroup$
8
  • $\begingroup$ So the point of the map being volume-preserving is that $\mathrm{vol}(M_+)=\mathrm{vol}(M_+)=\frac{1}{2}V_{\text{ball}},$ right? Also, I'm feeling a bit slow today - would you be so kind as to include the modification? Thank you. $\endgroup$ – Bobson Dugnutt Jun 12 '17 at 14:58
  • $\begingroup$ I meant $\mathrm{vol}(M_+)=\mathrm{vol}(M_-)$, of course. $\endgroup$ – Bobson Dugnutt Jun 12 '17 at 15:20
  • $\begingroup$ Yes, that's right, they have the same volume, therefore both are $\frac{1}{2} V_{\mathrm{ball}}$. In the same vein, we can define $9$ subsets according to which relations ($>, =, <$) hold for each of the inequalities. This time, five of them (precisely those involving $=$) will not contribute to the volume. Now, define the volume-preserving map $\psi : (x, y, z, w) \mapsto (-x, -y, z, w)$, and show that you can map the given region to any of the other three contributing subsets with appropriate compositions of $\phi, \psi$. $\endgroup$ – Travis Willse Jun 12 '17 at 16:07
  • $\begingroup$ Thanks, that was very clear! But I'm still not sure I understand why the subsets with an $=$ sign have zero volume. I guess it is like if we had a $3$-ball $x^2+y^2+z^2\leq 1$ .. if we now change the $\leq$ to an $=$, the resulting region is the unit $2$-sphere, which is two-dimensional and therefore does not contribute to the volume of the sphere. But if the ball is 4D, the sphere becomes 3D - surely that has a volume? I think my concept of "volume" is too tied to 3D structures, but where exactly am I going awry? $\endgroup$ – Bobson Dugnutt Jun 12 '17 at 16:54
  • $\begingroup$ That's right---we're measuring $4$-dimensional volume, and, roughly speaking, $3$-dimensional objects have zero $4$-dimensional volume, in analogy with your $3$- and $2$-dimensional example. $\endgroup$ – Travis Willse Jun 12 '17 at 16:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.